Question

Let N denote the set of all natural numbers. Define two binary relations on N as R₁ = {(x, y) ∈ N × N : 2x + y = 10} and R₂ = {(x, y) ∈ N × N : x + 2y = 10}. Then ______.

Options

• Both R₁ and R₂ are transitive relations

• Both R₁ and R₂ are symmetric relations

• Range of R₂ is {1, 2, 3, 4}

• Range of R₁ is {2, 4, 8}

Answer 1

Let N denote the set of all natural numbers. Define two binary relations on N as R₁ = {(x, y) ∈ N × N : 2x + y = 10} and R₂ = {(x, y) ∈ N × N : x + 2y = 10}. Then Range of R₂ is {1, 2, 3, 4}.

Explanation:

Here,

R₁ = {(x, y) ∈ N × N : 2x + y = 10} and 

R₂ = {(x, y) ∈ N × N : x + 2y = 10}

For R₁; 2x + y = 10 and x, y ∈ N

So, possible values for x and y are:

x = 1, y = 8 i.e. (1, 8);

x = 2, y = 6 i.e. (2, 6);

x = 3, y = 4 i.e. (3, 4) and

x = 4, y = 2 i.e. (4, 2);

R₁ = {(1, 8), (2, 6), (3, 4), (4, 2)}

Therefore, Range of R₁ is {2, 4, 6, 8}

R₁ is not symmetric.

Also, R₁ is not transitive because (3, 4), (4, 2) ∈ R₁ but (3, 2) ∉ R₁

Thus, options A, B and D are incorrect.

For R₂; x + 2y = 10 and x, y ∈ N

So, possible values for x and y are:

x = 8, y = 1 i.e. (8, 1);

x = 6, y = 2 i.e. (6, 2);

x = 4, y = 3 i.e. (4, 3) and

x = 2, y = 4 i.e. (2, 4)

R₂ = {(8, 1), (6, 2), (4, 3), (2, 4)}

Therefore, Range of R₂ is {1, 2, 3, 4}

R₂ is not symmetric and transitive.